Monty Hall Problem Revisited

After attempting to answer someone's supposed counterargument in the comments section to my earlier post, I realized that there's a better explanation of the Monty Hall Problem that will hopefully satisfy/silence the doubters. First, I'll restate the problem:

Monty Hall Problem

Of three doors, one is good. First, you select a door. Second, Monty Hall eliminates one of the doors for you - the door he eliminates is not the door you selected, and not the good door. Third, you have the option of keeping your original door, or changing to the only remaining door. What should you do?

SOLUTION

For the sake of this explanation, imagine that you play the game many times, and you always pick Door A, which means Monty will always have to eliminate either Door B or Door C. Every other aspect of the game is still fair and random, meaning that at the beginning of each game, Door A, Door B and Door C have an equal probability of being good.
Now let's consider the moment after you've selected Door A (because you always select Door A in this example) and Monty must eliminate Door B or Door C for you.

1. If Door B is good, he will eliminate Door C (1/3 of all occurences)
2. If Door C is good, he will eliminate Door B (1/3 of all occurences)
3. If Door A is good, Monty must choose whether to eliminate Door B or Door C. He does so with equal probability, so:

• If Door A is good, sometimes Monty eliminates Door C (1/6 of all occurences)
• If Door A is good, sometimes Monty eliminates Door B (1/6 of all occurences)

In 1/2 of all the games you play, Monty eliminates Door C (1/2 = 1/3 + 1/6) and you must then choose between Door A and Door B:
That means that for every 6 times you play the game, you'll have to choose between Door A and Door B 3 times. Of those 3 times, 2 of them will occur because Door B was the good door, and 1 of them will occur because Door A was the good door and Monty randomly eliminated Door C.

And the other 1/2 of the time (3 of every 6 games) you find yourself left with a choice between Door A and Door C, and it's twice as likely that Door C is the good door:
In either event, your chances are better if you change your original selection.